Double summation with improper integral

since $\frac{x^b}{e^{ax} b!}$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then: \begin{align} &\sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx \\ &=\int_{0}^\infty \sum_{a=2}^\infty e^{-ax} \sum_{b=1}^\infty \frac{x^{b}}{ \ b!} \ dx \\ &= \int_{0}^\infty \underbrace{\left(\sum_{a=2}^\infty (e^{-x})^a\right)}_{\text{geometric series}} \overbrace{\left(\sum_{b=0}^\infty \frac{x^{b}}{ \ b!}-1\right)}^{\text{series definition of $e^x$}} \ dx \\ &= \int_{0}^\infty \frac{1}{e^x(e^x-1)}(e^{x}-1)dx \\ &= \int_0^\infty e^{-x} dx \\&= 1.\end{align}

The integral is of the Gamma type,

$$\int_{0}^\infty \frac{x^{b}}{e^{ax}} \ dx=\frac1{a^{b+1}}\int_{0}^\infty t^be^{-t}\ dx =\frac{b!}{a^{b+1}}.$$

Then

$$\sum_{a=2}^\infty \sum_{b=1}^\infty\frac1{a^{b+1}}=\sum_{a=2}^\infty \frac1{a^2\left(1-\dfrac1a\right)}=\sum_{a=2}^\infty \frac1{a(a-1)}$$ is indeed a telescoping sum, giving

$$\frac1{2-1}.$$

For the first part I often use the Laplace transform: $$\frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx = \frac{1}{b!} \int_0^\infty x^be^{-ax}\ dx = \frac{1}{b!} {\cal L}(x^b)\Big|_{s=a} = \frac{1}{b!} \frac{b!}{a^{b+1}} = \frac{1}{a^{b+1}}$$ this make it easier.